It is possible your sensor uses Hart communications, which is a FSK modulated digital signal on top of the 4-20mA analogue signal.
The basic issue here is that the sensor has a huge range, so if you use analogue signal processing to convert the signal it is likely that errors will come in. While the analogue processing can be made reasonably accurate by using better quality amplifiers and precision references and precision resistors (low temperature coefficient = 10ppm) the original signal when expanded may not be so accurate. For example, is the temperature with 4mA exactly -200C. How do you even test that? Reasonable accuracy to expect is 1/4000 of 800 degrees = 0.2 degrees, in the analogue processing alone.
If what you say is true, -200°C is 4mA, 0°C is 8mA and one hundred degrees is represented by 10mA. Note that 4mA is zero, the bottom of the scale, not zero degrees. The scale is 800°C over 16mA = 0.02mA per degree.
Use a resistor in the 4-20mA line to convert to voltage. With 200 ohms, then the 4-20mA becomes 0.8V to 4V. The voltage is 4mV/°C. An instrumentation amplifier (INA) converts this voltage across the resistor which is at some unknown potential referred to the amplifier ground, to a voltage referenced to the ground pin of the amplifier. See the first link. The gain is assumed to be exactly one in the following. The 2 differential inputs of this amplifier connect across the resistor. I would use 2.2K protective resistors in series with the inputs. The reference (ground) input of the amplifier could be offset by a buffered -0.8V (the ground reference is a low impedance input) so the amplifier output is 0-3.2V voltage analogue for -200 to +600 degrees. This removes the 4mA offset. A suitable device is the LT1167.
Assuming you want 0V to be 0 degrees, this requires a further offset in the same direction. We can say that instead of 4mA being zero volts, 8mA=0°C will be zero volts. The required offset is therefore 1.6V exactly, from a precision reference, and buffered by an opamp like the LT1012, so it can drive the reference ground accurately. The input range of 0-100°C now represents a voltage from 0-400mV. This drives the voltage to current converter. An intermediate stage can provide additional gain, for example a gain of 10 would provide 40mV per degree, a 0-4V range. Otherwise work out the gain and new offset for the INA.
The fourth and fifth links refer to a voltage to 4-20mA current loop converter. There are also specific devices (ICs) that work as 4-20mA converters with a voltage input. The reference for 4mA offset in the 4-20mA converter can be eliminated by a further adjustment to the offset of the voltage offset along the line.
The reason I suggest the Linear Technology devices is because these can be simulated using LTspice, which can be downloaded free from the LT website. It will pay to make sure it works. Get the polarities and gains and offsets right before your start.